Find dy/dx of square root of seven plus square root of three x plus one

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \sqrt{7 + \sqrt{3x + 1}} \), we will apply the chain rule.

Given:
\[
y = \sqrt{7 + \sqrt{3x + 1}}
\]

### Step 1: Differentiate the outer square root function
Let \( u = 7 + \sqrt{3x + 1} \), so:
\[
y = \sqrt{u}
\]
The derivative of \( y \) with respect to \( u \) is:
\[
\frac{dy}{du} = \frac{1}{2\sqrt{u}}
\]

### Step 2: Differentiate the inner function \( u = 7 + \sqrt{3x + 1} \)
Let \( v = 3x + 1 \), so:
\[
u = 7 + \sqrt{v}
\]
The derivative of \( u \) with respect to \( v \) is:
\[
\frac{du}{dv} = \frac{1}{2\sqrt{v}}
\]

### Step 3: Differentiate the innermost function \( v = 3x + 1 \)
The derivative of \( v \) with respect to \( x \) is:
\[
\frac{dv}{dx} = 3
\]

### Step 4: Combine all derivatives using the chain rule
Now, applying the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}
\]
Substitute the derivatives:
\[
\frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{1}{2\sqrt{v}} \cdot 3
\]

### Step 5: Substitute back the expressions for \( u \) and \( v \)
\[
\frac{dy}{dx} = \frac{3}{4\sqrt{7 + \sqrt{3x + 1}} \cdot \sqrt{3x + 1}}
\]

So, the derivative is:
\[
\frac{dy}{dx} = \frac{3}{4\sqrt{7 + \sqrt{3x + 1}} \cdot \sqrt{3x + 1}}
\]

Would you like more details on any of these steps or do you have any questions?

### Related Questions:
1. How do you apply the chain rule to more complex functions?
2. What is the chain rule and how does it work?
3. How can we simplify derivatives of nested square roots?
4. What is the importance of correctly identifying inner and outer functions in the chain rule?
5. How can you verify your derivative calculation?

### Tip:
Always clearly identify the inner and outer functions when using the chain rule to avoid confusion during differentiation.

Discover how to find the derivative dy/dx of sqrt(7 + sqrt(3x + 1)) with a detailed step-by-step solution using the chain rule. This problem involves differentiation of nested square roots, suitable for advanced high school students and calculus enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation